# Is the spring constant k changed when you divide a spring into parts?

I've always been taught that the spring constant $k$ is a constant — that is, for a given spring, $k$ will always be the same, regardless of what you do to the spring.

My friend's physics professor gave a practice problem in which a spring of length $L$ was cut into four parts of length $L/4$. He claimed that the spring constant in each of the new springs cut from the old spring ($k_\text{new}$) was therefore equal to $k_\text{orig}/4$.

Is this true? Every person I've asked seems to think that this is false, and that $k$ will be the same even if you cut the spring into parts. Is there a good explanation of whether $k$ will be the same after cutting the spring or not? It seems like if it's an inherent property of the spring it shouldn't change, so if it does, why?

• It is $k_{\rm new}=4k_{\rm orig}$, not $k_{\rm new}=k_{\rm orig}/4$, see springs in series e.g. on Wikipedia. – Qmechanic Nov 16 '11 at 21:09
• Note that the spring constant $k = \frac{EA_0}{L_0}$ where E is the Young's modulus of the material (constant, at least in one direction), $A_0$ is the unstretched cross sectional area, and $L_0$ is the unstretched length of the ideal spring. This indicates that $k$ is inversely proportional to the original length of the spring, with all else (material, cross section) being same. Therefore a spring a quarter of the length will have a spring constant four times as high. – Asad Saeeduddin Nov 27 '13 at 5:30

Well, the sentence

It seems like if it's an inherent property of the spring it shouldn't change, so if it does, why?

clearly isn't a valid argument to calculate the $k$ of the smaller springs. They're different springs than their large parent so they may have different values of an "inherent property": if a pizza is divided to 4 smaller pieces, the inherent property "mass" of the smaller pizzas is also different than the mass of the large one. ;-)

You may have meant that it is an "intensive" property (like a density or temperature) which wouldn't change after the cutting of a big spring, but you have offered no evidence that it's "intensive" in this sense. No surprise, this statement is incorrect as I'm going to show.

One may calculate the right answer in many ways. For example, we may consider the energy of the spring. It is equal to $k_{\rm big}x_{\rm big}^2/2$ where $x_{\rm big}$ is the deviation (distance) from the equilibrium position. We may also imagine that the big spring is a collection of 4 equal smaller strings attached to each other.

In this picture, each of the 4 springs has the deviation $x_{\rm small} = x_{\rm big}/4$ and the energy of each spring is $$E_{\rm small} = \frac{1}{2} k_{\rm small} x_{\rm small}^2 = \frac{1}{2} k_{\rm small} \frac{x_{\rm big}^2}{16}$$ Because we have 4 such small springs, the total energy is $$E_{\rm 4 \,small} = \frac{1}{2} k_{\rm small} \frac{x_{\rm big}^2}{4}$$ That must be equal to the potential energy of the single big spring because it's the same object $$= E_{\rm big} = \frac{1}{2} k_{\rm big} x_{\rm big}^2$$ which implies, after you divide the same factors on both sides, $$k_{\rm big} = \frac{k_{\rm small}}{4}$$ So the spring constant of the smaller springs is actually 4 times larger than the spring constant of the big spring.

You could get the same result via forces, too. The large spring has some forces $F=k_{\rm big}x_{\rm big}$ on both ends. When you divide it to four small springs, there are still the same forces $\pm F$ on each boundary of the smaller strings. They must be equal to $F=k_{\rm small} x_{\rm small}$ because the same formula holds for the smaller springs as well. Because $x_{\rm small} = x_{\rm big}/4$, you see that $k_{\rm small} = 4k_{\rm big}$. It's harder to change the length of the shorter spring because it's short to start with, so you need a 4 times larger force which is why the spring constant of the small spring is 4 times higher.

• It's equivalent if I understand you well. – Luboš Motl Apr 4 '17 at 12:41

To supplement the answer by Luboš Motl, I will come to this problem from a Material Science point of view.

What you mean by the inherent property of the string is not the spring constant, in fact, it is Young's modulus $E$, which only depends on the properties of a material of a body but not it's shape.

$$E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{\sigma}{\varepsilon} = \frac{\text{force per area}}{\text{extension per length}} = \frac{F / A}{x / l} = \frac{F l }{x A}$$

Now use this definition to construct the Hooke's Law: $$F = \frac{EA}{l} x = k x$$ where we see that $$k = \frac{EA}{l}$$

Now consider what happens when we divide the spring. We change only the length of the spring, whilst keeping A (same cross-section area) and E (same spring, same material) the same. When we make the spring four times shorter we essentially have the following:

$$k_\text{old} = \frac{EA}{l_\text{old}} = \frac{EA}{4 l_\text{new}} = \frac{1}{4} \frac{EA}{l_\text{new}} = \frac{1}{4} k_\text{new}$$

Note, that this is assuming a rubber band like set-up, where we assume that the spring can be modelled by a uniform bar of elastic material. A more rigorous proof of the dependence of spring constant and the length of the spring would involve the geometry of the spring and various torques on the spring elements when it is under load. However, all this complication just brings additional pre-factors to the spring constant, which are independent of the length of the spring.

# A heuristic derivation of the Young Modulus-Force relation

I thought I might talk about why $E$ is always constant for some type of material. All bonds between atoms can be thought as tiny springs obeying Hooke's law in case of small displacements.

Because of the energy conservation we already know (the answer by Luboš Motl), that if we connect several springs, then we will change the effective spring constant: $$k_\text{new} = k / n$$ where $n$ is the number of the springs and $k$ is the single bond spring constant.

Hence, for the same extension, the force scales with the length of the spring as follows: $$F = \frac{k x}{n} = k\frac{l_\text{unit}}{l}x = kxl_\text{unit} \times \frac{1}{l} = \text{const.} \times \frac{x}{l}$$

Now, what about connecting the strings in parallel? From the energy conservation argument, we know, that the effective spring constant then will change in a different way: $$k_\text{new} = kn$$ where $n$ now will be related to the surface area of the material.

Now, the force for the same extension scales as: $$F = knx = k\rho x \times A = \text{const.} \times Ax$$ where $\rho$ is the density of the springs.

There are only two ways of combining strings (in parallel or in series), hence the overall formula for force must be of a form bellow:

$$F = E \times \frac{A}{l} x$$

And we can call that unknown constant $E$ the Young's modulus, which we know will be specific to the material (i.e. the nature of those chemical bonds). What is more, because of our analysis above, we know that for a given material the remaining unknown quantity $E$ will be independent of the cross-sectional area, length or extension of the spring.

So with a very simple thinking and some basic knowledge of the energy conservation, we could recover the law I assumed in my first part of explanation.

EDIT: I noticed that there were some errors in the second part of my explanation, hence a complete overhaul. Also, I hope I clarified the first part of the explanation.

• When you say "but you change l by a factor of 4. This makes the new k to be 4 times smaller" To make the wording similar to OP's question I suppose you mean "if $l$ is cut by a factor of 4 then $k$ increases by a factor of 4." – Maesumi Feb 5 '14 at 18:18
• I amended my explanation, hope it makes more sense now. – aignas Feb 7 '14 at 13:56
• @gns-ank What do you mean by density of the springs? $\rho = \frac{n}{A}$? If that's the case you should change $\frac{A}{\rho}$ to $A \times \rho$. Also, could you explain your last step, i.e., how to obtain the overall formula $F = E \times \frac{A}{l} x$? – 21Brunoh Nov 1 '14 at 1:39
• @21Brunoh, yes, by 'density of the springs' I meant that $\rho = n/A$ and there was a typo in the answer. Should be fixed now. The last formula is obtained by the fact, that we derive two relationships between the force and the other quantities. In the first case we keep $A$ constant, in the other, we keep $l$ constant. Because we know, that there is no other arrangement of the springs, we can arrive at the final formula. – aignas Nov 4 '14 at 21:25

For a given spring, $k$ is a constant, As long as you're talking about an ideal spring. In other words, the definition of the ideal spring is that it applies the force proportional to its deformation length (at both endings of course).

I'm afraid both you and your professor are wrong. The correct formula should be:

$k_{new} = k_{orig}*4$

To show that let's do the following gedankenexperiment. Suppose you have your original spring in tension. It's deformed length $L$, and applies the appropriate force $F$.

Now imagine that your spring is actually 4 consequently connected springs of length $L/4$. Each spring is at rest, this means that for every spring the forces applied to both endings are equal. Since all the springs are connected and apply forces on each other - this means that all the forces applied to all the spring endings are the same. And obviously they equal to $F$.

OTOH each spring is deformed by only $L/4$. Hence - their "constants" are 4 times higher

In other way $k \times l = \text{constant}$ so $K$ varies $1/l$.

So $K\times l=K' \times l/4$ and $K=K'/4$ so $K'=4K$ $K'$ will be $4k$ ($k$=original spring constant)

In problems like this, it's easier to work with a specific case as opposed to the general one. Let's assume that the original spring has 12 coils. Then each of the cut springs would have 3 coils.

Let's say that applying a 12 N force to the original spring would compress it 12 cm. Now let's apply some force, Y, to the spring that was cut (has 3 coils). This force compresses the cut spring by 12 cm as well.

Here, our intuition tells us that Y should be much greater than 12 N since compressing a small spring by 12 cm should require more force than compressing a large spring by 12 cm. So how does this work and what is this force, Y?

Essentially we are compressing each coil by 1 cm in the original spring but we are compressing each coil by 4 cm in the cut spring.

Force, however, is transmitted across the entire spring, something that we frequently forget (we also forget this a lot when looking at a rope under tension and that the tension is the same at each at some point on the rope). So the force on the spring is the same at every point on the spring.

This means that 12 N of force is being applied to each of the coils in the original spring to compress each coil by 1 cm. In other words, 12 N of force are required to compress 12 coils by 12 cm in total and also to compress 1 coil by 1 cm in total. Yes, that's right, applying 12 N will cause 12 coils to compress by 12 cm but applying 1 N to 1 coil will not compress the coil by 1 cm; 12 N will compress 1 coil by 1 cm (this does not seem intuitive so let me rephrase it this way: 12 N compresses the original spring by 12 cm and one coil of the spring by 1 cm).

Then 48 N of force are required to compress 12 coils by 48 cm in total and also to compress 1 coil by 4 cm in total. Using this logic, 48 N of force would be required to compress 3 coils by 12 cm in total. So Y=48 N. When we disregard direction, the equation F=-kx becomes F=kx. So k=F/x. Plugging in the values from the original spring gives k=1 N/cm. Plugging in the values from the cut spring gives k=4 N/cm. So yes, your teacher was wrong because his logic would give 1/4 N/cm not 4 N/cm but your friends were wrong too because k is not constant regardless of the length of the spring. And it makes sense intuitively: it's harder to compress a short spring by some distance than it is to compress a longer spring by that same distance.